If the hydrogen ion concentration is 10-7M, what is the pH of the solution? Show all work.

1. In the first step of the mechanism for this process, a phenoxide anion is generated. This phenoxide anion goes on to act as a

Tju [1.3M]

Answer:

See the explanation

Explanation:

In this case, in order to get an elimination reaction we need to have a strong base. In this case, the base is the phenoxide ion produced the phenol (see figure 1).

Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).

Finally, the phenoxide will attack the primary carbon attached to the Cl. The C-Cl bond would be broken and the C-O would be produced at the same time to get a substitution (see figure 1).

3 0

3 years ago

please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi

Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -Assuming STP-. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

Volume Methane:

3600s * (0.25L / s) = 900L Methane

Moles methane:

PV = nRT; PV / RT = n

Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

Mass methane:

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>

5 0

2 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

Thus, percentage abundance of 121 Sb is = 57.2 %

percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %

6 0

2 years ago

Which pair of formulas represents the empirical formula and the molecular formula of a compound?

AnnyKZ [126]

Answer:

CH2O and C6H12O6

Explanation:

To find an empirical formula, take a molecular formula and divide the subscript of each element by the greatest common factor of all the subscripts. In this case, the only pair that works is CH2O,C6H12O6, which can be verified by dividing the coefficients of the molecular formula by 6.

3 0

2 years ago

How many grams of aluminum sulfide would form if 2.25 liters of a 1.50 molar aluminum chloride solution reacts with 2.00 liters

Alex787 [66]

Hope this helps you!

8 0

2 years ago